Question

In this section I've formulated the cluster expansion for a gas with a fixed number of particles, using the "canonical" formalism of Chapter 6. A somewhat cleaner approach, however, is to use the "grand canonical" formalism introduced in Section 7.1, in which we allow the system to exchange particles with a much larger reservoir.

(a) Write down a formula for the grand partition function (Z) of a weakly interacting gas in thermal and diffusive equilibrium with a reservoir at fixed T andµ. Express Z as a sum over all possible particle numbers N, with each term involving the ordinary partition function Z(N).

(b) Use equations 8.6 and 8.20 to express Z(N) as a sum of diagrams, then carry out the sum over N, diagram by diagram. Express the result as a sum of similar diagrams, but with a new rule 1 that associates the expression (>./vQ) J d3ri with each dot, where >. = e13µ,. Now, with the awkward factors of N(N - 1) · · · taken care of, you should find that the sum of all diagrams organizes itself into exponential form, resulting in the formula

Note that the exponent contains all connected diagrams, including those that can be disconnected by removal of a single line.

(c) Using the properties of the grand partition function (see Problem 7.7), find diagrammatic expressions for the average number of particles and the pressure of this gas.

(d) Keeping only the first diagram in each sum, express N(µ) and P(µ) in terms of an integral of the Mayer /-function. Eliminate µ to obtain the same result for the pressure (and the second virial coefficient) as derived in the text.

(e) Repeat part (d) keeping the three-dot diagrams as well, to obtain an expression for the third virial coefficient in terms of an integral of /-functions. You should find that the A-shaped diagram cancels, leaving only the triangle diagram to contribute to C(T).

Short answer

The required equation is$Z=\frac{1}{N!}·(\sqrt{2\pi mkT}{)}^3·\int d^3{r}_1\cdots d^3{r}_N·e^{-\beta U_{pot}}$.

Step-by-step solution

Given information

We have been given that $Z_1=\frac{1}{h^3}·\int d^{3}r{d}^{3}p·e^{-\beta U}$

Simplify

For same molecules

$Z_1=\frac{1}{h^3}·\int d^{3}r{d}^{3}p·e^{-\beta U}$

Energy can be written as:

$U=\frac{{\left|\overrightarrow{p_1}\right|}^2}{2m}+\frac{{\left|\overrightarrow{p_2}\right|}^2}{2m}+\dots +\frac{{\left|\overrightarrow{p_N}\right|}^2}{2m}$

$Z=\frac{1}{N!}·\frac{1}{h^{3N}}·\int d^3{r}_1\cdots d^3{r}_N{d}^3{p}_1\cdots d^3{p}_N·e^{-\frac{\beta ·{\left|{\overrightarrow{p}}_1\right|}^2}{2m}}\cdots e^{-\frac{\beta ·{\left|\overrightarrow{p_N}\right|}^2}{2m}}·e^{-\beta U_{ppt}}$

We can write virial equation

$=2\pi ·{\int }_0^{\infty }{r}^2\left(e^{-\beta U\left(r\right)}-1\right)dr$